$\begin{aligned} &g(x)=-20-3x \\\\ &h(x)=2-\left(\dfrac{1}{2}\right)^x \end{aligned}$ $(g\circ h) (-2)=$
Solution: Let's start by rewriting $(g\circ h) (-2)$ as $g(h(-2))$. When evaluating composite functions, we work our way inside out. To evaluate $g(h(-2))$, let's first evaluate $h(-2)$. Then we'll plug that result into $g$ to find our answer. Let's evaluate $h({-2})$. $\begin{aligned}h(x)&=2-\left(\dfrac{1}{2}\right)^x\\\\ h({-2})&=2-\left(\dfrac{1}{2}\right)^{{-2}}~~~~~~~~~~\text{Plug in }x={-2}\\\\ &=2-4\\\\ &={-2}\end{aligned}$ We now know that $g(h({-2}))$ is the same as $g({-2})$ because $h({-2}) = {-2}$. Let's evaluate $g({-2})$. $\begin{aligned}g(x)&=-20-3x\\\\ g({{-2}})&=-20-3({-2})~~~~~~~~~~\text{Plug in }x={-2}\\\\ &=-20+6\\\\ &=-14\end{aligned}$ The answer: $(g\circ h)(-2) =-14$